What is int ( 2 + x ) / sqrt ( 4 - 2x - x^2 dx?

1 Answer

int (2+x)/sqrt(4-2x-x^2)dx=arcsin ((x+1)/sqrt5)-sqrt(4-2x-x^2)+C

Explanation:

Start from the given

int (2+x)/sqrt(4-2x-x^2)dx

Start with Algebra by completing the square

4-2x-x^2=-(x^2+2x-4)=-(x^2+2x+1-1-4)

and

4-2x-x^2=-((x+1)^2-5)=5-(x+1)^2

then

int (2+x)/sqrt(4-2x-x^2)dx=int (2+x)/sqrt(5-(x+1)^2)dx

The Trigonometric Substitution

Let x+1=sqrt(5)*sin theta
and x=sqrt(5)*sin theta -1
and dx=sqrt(5)*cos d theta

Let's do the substitution

int (2+x)/sqrt(5-(x+1)^2)dx=
int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/sqrt(5-(sqrt(5)*sin theta)^2)

int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/sqrt(5-5*sin^2 theta)

continue simplification by trigonometric identities

int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*sqrt(1-sin^2 theta))

int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*sqrt(cos^2 theta))

int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*cos theta)

int((sqrt(5)sin theta + 1)*(sqrt(5)*cos theta)d theta)/(sqrt5*cos theta)

and

int (2+x)/sqrt(5-(x+1)^2)dx=int(sqrt(5)sin theta + 1)d theta

int (2+x)/sqrt(5-(x+1)^2)dx=int(1+sqrt(5)sin theta )d theta

int (2+x)/sqrt(5-(x+1)^2)dx=theta-sqrt(5)*cos theta+C

Now, time to imagine your right triangle with
angle theta
Let x+1 the Opposite side to angle theta
Let sqrt5 the Hypotenuse
Let sqrt(4-2x-x^2) the Adjacent side to angle theta

Return the variables

int (2+x)/sqrt(5-(x+1)^2)dx=theta-sqrt(5)*cos theta+C

int (2+x)/sqrt(5-(x+1)^2)dx=arcsin ((x+1)/(sqrt5))-sqrt(4-2x-x^2)+C

I hope the explanation is useful....God bless...